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MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 9 10/2/2013

Conditional expectations, filtration and martingales

Content.

1. Conditional expectations 2. Martingales, sub-martingales and super-martingales

1 Conditional Expectations

1.1 Definition

Recall how we define conditional expectations. Given a X and an event A we define [X|A] = E[X1{A}] . E P(A) Also we can consider conditional expectations with respect to random vari­ ables. For simplicity say Y is a simple random variable on Ω taking values y1, y2, . . . , yn with some

P(ω : Y (ω) = yi) = .

Now we define conditional expectation E[X|Y ] as a random variable which takes value E[X|Y = yi] with pi, where E[X|Y = yi] should be understood as expectation of X conditioned on the event {ω ∈ Ω: Y (ω) = yi}. It turns out that one can define conditional expectation with respect to a σ­ field. This notion will include both conditioning on events and conditioning on random variables as special cases.

Definition 1. Given Ω, two σ-fields G ⊂ F on Ω, and a probability P on (Ω, F). Suppose X is a random variable with respect to F but not necessar­ ily with respect to G, and suppose X has a finite L1 norm (that is E[|X|] < ∞). The conditional expectation E[X|G] is defined to be a random variable Y which satisfies the following properties:

(a) Y is measurable with respect to G.

1 (b) For every A ∈ G, we have E[X1{A}] = E[Y 1{A}]. For simplicity, from now on we write Z ∈ F to indicate that Z is measurable with respect to F. Also let F(Z) denote the smallest σ-field such with respect to which Z is measurable.

Theorem 1. The conditional expectation E[X|G] exists and is unique. Uniqueness that if Y ' ∈ G is any other random variable satisfying ' conditions (a),(b), then Y = Y a.s. (with respect to measure P). We will prove this theorem using the notion of Radon-Nikodym , the existence of which we state without a proof below. But before we do this, let us develop some intuition behind this definition.

1.2 Simple properties

• Consider the trivial case when G = {Ø, Ω}. We claim that the constant value c = E[X] is E[X|G]. Indeed, any constant is measurable with respect to any σ-field So (a) holds. For (b), we have E[X1{Ω}] = E[X] = c and E[c1{Ω}] = E[c] = c; and E[X1{Ø}] = 0 and E[c1{Ø}] = 0.

• As the other extreme, suppose G = F. Then we claim that X = E[X|G]. The condition (b) trivially holds. The condition (a) also holds because of the equality between two σ-fields.

• Let us go back to our example of conditional expectation with respect to an event A ⊂ Ω. Consider the associated σ-fields G = {Ø, A, Ac , Ω} (we established in the first lecture that this is indeed a σ-field). Consider a random variable Y :Ω → R defined as E[X1{A}] Y (ω) = E[X|A] = c1 P(A) for ω ∈ A and c c E[X1{A }] Y (ω) = E[X|A ] = c2 P(Ac) c for ω ∈ A . We claim that Y = E[X|G]. First Y ∈ G. Indeed, assume for simplicity c1 < c2. Then {ω : Y (ω) ≤ x} = Ø when x < c1, = A

2 for c1 ≤ x < c2, = Ω when x ≥ c2. Thus Y ∈ G. Then we need to check c equality E[X1{B}] = E[Y 1{B}] for every B = Ø, A, A , Ω, which is straightforward to do. For example say B = A. Then

E[X1{A}] = E[X|A]P(A) = c1P(A).

On the other hand we defined Y (ω) = c1 for all ω ∈ A. Thus

E[Y 1{A}] = c1E[1{A}] = c1P(A). And the equality checks.

• Suppose now G corresponds to some partition A1,...,Am of the sample Ω. Given X ∈ F, using a similar analysis, we can check that Y = E[X|G] is a random variable which takes values E[X|Aj ] for all ω ∈ Aj , for j = 1, 2, . . . , m. You will recognize that this is one of our earlier examples where we considered conditioning on a simple random variable Y to get E[X|Y ]. In fact this generalizes as follows:

• Given two random variables X, Y :Ω → R, suppose both ∈ F. Let G = G(Y ) ⊂ F be the field generated by Y . We define E[X|Y ] to be E[X|G].

1.3 Proof of existence We now give a proof sketch of Theorem 1.

Proof. Given two probability measures P1, P2 defined on the same (Ω, F), P2 is defined to be absolutely continuous with respect to P1 if for every A ∈ F, P1(A) = 0 implies P2(A) = 0. The following theorem is the main technical part for our proof. It involves using the familiar idea of change of measures.

Theorem 2 (Radon-Nikodym Theorem). Suppose P2 is absolutely continuous with respect to P1. Then there exists a non-negative random variable Y :Ω → R+ such that for every A ∈ F

P2(A) = EP1 [Y 1{A}].

Function Y is called Radon-Nikodym (RN) derivative and sometimes is denoted dP2/dP1.

3 Problem 1. Prove that Y is unique up-to measure zero. That is if Y ' is also RN ' derivative, then Y = Y a.s. w.r.t. P1 and hence P2. We now use this theorem to establish the existence of conditional expec­ tations. Thus we have G ⊂ F, P is a on F and X is measurable with respect to F. We will only consider the case X ≥ 0 such that E[X] < ∞. We also assume that X is not constant, so that E[X] > 0. Consider a new probability measure P2 on G defined as follows:

EP[X1{A}] P2(A) = ,A ∈ G, EP[X] where we write EP in place of E to emphasize that the expectation operator is with respect to the original measure P. Check that this is indeed a probability measure on (Ω, G). Now P also induced a probability measure on (Ω, G). We claim that P2 is absolutely continuous with respect to P. Indeed if P(A) = 0 then the numerator is zero. By the Radon-Nikodym Theorem then there exists Z which is measurable with respect to G such that for any A ∈ G

P2(A) = EP[Z1{A}].

We now take Y = ZEP[X]. Then Y satisfies the condition (b) of being a conditional expectation, since for every set B

EP[Y 1{B}] = EP[X]EP[Z1{B}] = EP[X1{B}]. The second part, corresponding to the uniqueness property is proved similarly to the uniqueness of the RN derivative (Problem 1).

2 Properties

Here are some additional properties of conditional expectations.

Linearity. E[aX + Y |G] = aE[X|G] + E[Y |G].

Monotonicity. If X1 ≤ X2 a.s, then E[X1|G] ≤ E[X2|G]. Proof idea is similar to the one you need to use for Problem 1.

Independence. Problem 2. Suppose X is independent from G. Namely, for every measurable A ⊂ R,B ∈ G P({X ∈ A} ∩ B) = P(X ∈ A)P(B). Prove that E[X|G] = E[X].

4 Conditional Jensen’s inequality. Let φ be a convex function and E[|X|], E[|φ(X)|] < ∞. Then φ(E[X|G]) ≤ E[φ(X)|G]. Proof. We use the following representation of a convex function, which we do not prove (see Durrett [1]). Let

A = {(a, b) ∈ Q : ax + b ≤ φ(x), ∀ x}.

Then φ(x) = sup{ax + b :(a, b) ∈ A}. Now we prove the Jensen’s inequality. For any pair of rationals a, b ∈ Q satisfying the bound above, we have, by monotonicity that E[φ(X)|G] ≥ aE[X|G] + b, a.s., implying E[φ(X)|G] ≥ sup{aE[X|G] + b :(a, b) ∈ A} = φ(E[X|G]) a.s.

Tower property. Suppose G1 ⊂ G2 ⊂ F. Then E[E[X|G1]|G2] = E[X|G1] and E[E[X|G2]|G1] = E[X|G1]. That is the smaller field wins.

Proof. By definition E[X|G1] is G1 measurable. Therefore it is G2 measurable. Then the first equality follows from the fact E[X|G] = X, when X ∈ G, which we established earlier. Now fix any A ∈ G1. Denote E[X|G1] by Y1 and E[X|G2] by Y2. Then Y1 ∈ G1,Y2 ∈ G2. Then

E[Y11{A}] = E[X1{A}], simply by the definition of Y1 = E[X|G1]. On the other hand, we also have A ∈ G2. Therefore

E[X1{A}] = E[Y21{A}].

Combining the two equalities we see that E[Y21{A}] = E[Y11{A}] for every A ∈ G1. Therefore, E[Y2|G1] = Y1, which is the desired result.

An important special case is when G1 is a trivial σ-field {Ø, Ω}. We obtain that for every field G

E[E[X|G]] = E[X].

5 3 and martingales

3.1 Definition

A family of σ-fields {Ft} is defined to be a filtration if Ft1 ⊂ Ft2 whenever t1 ≤ t2. We will consider only two cases when t ∈ Z+ or t ∈ R+. A process {Xt} is said to be adapted to filtration {Ft} if Xt ∈ Ft for every t.

Definition 2. A {Xt} adapted to a filtration {Ft} is defined to be a martingale if

1. E[|Xt|] < ∞ for all t.

2. E[Xt|Fs] = Xs, for all s < t. When equality is substituted with ≤, the process is called supermartingale. When it is substituted with ≥, the process is called submartingale.

Suppose we have a stochastic process {Xt} adapted to filtration {Ft} and ' suppose for some s < s < t we have E[Xt|Fs] = Xs and E[Xs|Fs/ ] = Xs/ . Then using Tower property of conditional expectations

E[Xt|Fs/ ] = E[E[Xt|Fs]|Fs/ ] = E[Xs|Fs/ ] = Xs/ .

This means that when the stochastic process {Xn} is discrete time it suffices to check E[Xn+1|Fn] = Xn for all n in order to make sure that it is a martingale.

3.2 Simple examples

1. . Let Xn, n = 1,... be an i.i.d. with µ 2 n and variance σ < ∞. Let Fn be the Borel σ-algebra on R . Then i Sn − µn = 0≤k≤n Xk − µn is a martingale. Indeed Sn is adapted to Fn, and

E[Sn+1 − (n + 1)µ|Fn] = E[Xn+1 − µ + Sn − nµ|Fn] = E[Xn+1 − µ|Fn] + E[Sn − nµ|Fn] a = E[Xn+1 − µ] + Sn − nµ

= Sn − nµ.

Here in (a) we used the fact that Xn+1 is independent from Fn and Sn ∈ Fn. 2. Random walk squared. Under the same setting, suppose in addition 2 2 µ = 0. Then Sn − nσ is a martingale. The proof of this fact is very similar.

6 4 Additional reading materials

• Durrett [1] Section 4.1, 4.2.

References

[1] R. Durrett, Probability: theory and examples, Duxbury Press, second edi­ tion, 1996.

7 MIT OpenCourseWare http://ocw.mit.edu

15.070J / 6.265J Advanced Stochastic Processes Fall 2013

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